#4) Symmedians in a Triangle

by AnArtist, Feb 7, 2018, 4:32 PM

Let $ABC$ be a triangle.

Let $T$ be a point inside the triangle such that the lines $AT, BT, CT$ are the $T$ symmedians of triangles $\Delta BTC , \Delta CTA, \Delta ATB$ respectively.

Let $D, E, F$ be the points on rays $AT, BT, CT$ such that $TD = 2AT, TE = 2BT, TF = 2CT$ .

Then prove that $T$ is the radical center of circles $(BDC), (CEA),(AFB)$ .

Source: Own

Solution in the Comments.

This post has been edited 4 times. Last edited by pro_4_ever, Mar 1, 2018, 10:40 AM

symmedian

radical axis

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Here is my solution-
Let $AT \cap (BTC) = D'$ where $D'$ is different from $T$ . Define $E' , F'$ similarly.

Now perform an inversion around $T$ with radius $r$ . Let $P^*$ denote the image of point $P$ under the inversion.

Claim 1) $T$ is the centroid of $\Delta A^*B^*C^*$
Proof-

Clearly $A^*T$ meets $B^*C^*$ at $D'^*$ .

Therefore it suffices to show that $D'^*$ is the midpoint of $B^*C^*$ .

By inversion distance formula.

$B^*D'^* = \frac{r^2. BD'}{TB.TD'}$
$C^*D'^* = \frac{r^2.CD'}{TC.TD'}$
Hence $\frac{B^*D'^*}{C^*D'^*}= \frac{BD'.TC}{BT.CD'}$
But since the quadrilateral $BTCD'$ is harmonic our claim is proved.

Claim 2) $D = D'$ .
Proof-

It suffices to prove,

$TD' = 2TA$
Note that,

$\frac{TD'}{TA} = \frac{\frac{r^2}{T^*D'^*}}{\frac{r^2}{T^*A^*}} = \frac{T^*A^*}{T^*D'^*}= 2$
Where the last conclusion follows from claim 1).

So $D' = D$ as claimed.

Now we move onto the problem.

By claim 2) $BTCD$ is cyclic. Similarly $ATCE$ and $BTAF$ is cyclic.

Hence we are done.

by AnArtist, Feb 8, 2018, 3:35 AM

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Solution by TelvCohl

TelvCohl wrote:

Let $\triangle XYZ$ be the antipedal triangle of $T$ WRT $\triangle ABC,$ then $X(T, \parallel YZ; Y, Z) = T(\perp TX, A; C,B) = -1,$ so $XT$ is the X-median of $\triangle XYZ.$ Analogously, we can prove $YT, ZT$ is the Y-median, Z-median of $\triangle XYZ,$ respectively, so $T$ is the centroid of $\triangle XYZ$ and $D \in \odot (BTC),$ $E \in \odot (CTA),$ $F \in \odot (ATB).$

Remark : $T$ is one of the foci of the Steiner inellipse of $\triangle ABC.$

I don't understand anything though.

by AnArtist, Feb 8, 2018, 5:10 AM

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What about my solution?

Clearly, the problem is done if we show the following:

$DBC$ is a triangle with the $D-$ symmedian meeting $(DBC)$ again at $T$ . The $D-$ Dumpty point of $DBC$ is $X$ (the midpoint of $DT$ ). Reflection of $X$ in $T$ is $A$ . Then $AT, CT$ are the symmedians in $BTC, BTA$ .

Proof: The first claim is trivial, since $BCTD$ is harmonic, so we focus on the second part. We need $\frac{\sin \angle BTC}{\sin \angle CTA} = \frac{BT}{TA}$ . It is well known as a property of the $D-$ Dumpty point that $(DXB)$ is tangent to $DC$ . So from angle chase, we have $\angle BXT = \angle D$ , and $\angle XBT = \angle B$ . So by the definition of $A$ , we are done, as $\angle BTC = 180 - \angle D, \angle CTA = 180 - \angle B$ .

Here I now show how the problem follows from the claim.
$AT/CT = \sin \angle ATB / \sin \angle BTC,$ so as $A$ varies on $DT$ extended, the length $AT$ is fixed. So $A$ is the unique point satisfying the first condition of the problem. Hence, $A=A'$ .

Remark : The $D-$ Dumpty point is also known as the vertex of the second Brocard triangle, and here.

It had a small detail left on the original thread, i.e. proving $A=A'$ (in the notation of the discussion of the thread that ensued) which was completed. (I thought that it was obvious from the sine rule chase, that the point $A$ had to be unique)

by WizardMath, Feb 9, 2018, 1:28 PM

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@WizardMath I'm so sorry I forgot to add your solution.

by AnArtist, Feb 9, 2018, 4:19 PM

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Hey, what a nice blog!!!!!
by Power_Set, Jun 24, 2018, 9:53 PM
Is your blog dead ? No new post in almost 2 months.

My blog has died due to malnutrition of problems.
by integrated_JRC, May 2, 2018, 5:52 AM
@ below, I will do it soon
by pro_4_ever, Mar 1, 2018, 10:37 AM
Hi, I think we should 'hide' our solutions, as it is difficult to go up and down the blog
by Drunken_Master, Mar 1, 2018, 10:36 AM
Yeah, I created a test blog, it seems to be an AoPS problem not CSS, can't do anything about it. I'll post it in site support forum later.
by Vrangr, Feb 20, 2018, 5:48 AM
The problem is not in the CSS, I believe, as the CSS only looks after the appearance of the blog, and not the $\text{\LaTeX}$ rendering.
by WizardMath, Feb 19, 2018, 11:34 PM
I'll fix the CSS after boards of need be (I'll probably be starting my own blog too then)
by Vrangr, Feb 19, 2018, 4:41 PM
@ below,
Pls ask WizardMath...
The CSS is taken from his amazing Blog, "An Olympiad Journey"...
I know nothing abt programming.
(BTW I got permission for taking the CSS. No Copyright Issues!)
by pro_4_ever, Feb 19, 2018, 4:16 PM
Huge flaw in website design: comment gets erased if while typing $\LaTeX$ has an error and needs to be fixed.
by Vrangr, Feb 19, 2018, 12:26 PM
Upon WizardMath's Request, let us all post synthetic solutions. Well, let's declare that this blog is intended to give the reader some good problems with simple synthetic solutions. All the best. Not trying to be rude,but I might delete upcoming Bashes
by pro_4_ever, Feb 13, 2018, 4:50 PM
Here's an idea. Ban all bashes on this blog from now on. Also try posting your own previous geo solutions as storage (I hope this is allowed). This would make the quality of the blog better. Bashes are basically write-once, read-never things. Why bother?
by WizardMath, Feb 13, 2018, 4:33 PM
Somebody type Egmo 2013 Problem 1.
No bash,purely troll problem.
by QWERTYphysics, Feb 13, 2018, 4:11 PM
I hate having to see a pure geometry blog being swamped to death with bashes.

Obviously a bash is a super terrible idea. Don't do one.
Unless you are bad at synthetic.
by WizardMath, Feb 13, 2018, 2:58 PM
@ayan Done:)
by pro_4_ever, Feb 13, 2018, 1:38 PM
9th shout
by MEGAKNIGHT, Feb 13, 2018, 11:23 AM
@below Pls don't post a bash if you are thinking about posting a new entry!
by pro_4_ever, Feb 13, 2018, 11:11 AM
Wait, I thought bash is frowned upon in this blog.
by WizardMath, Feb 12, 2018, 6:57 PM
Hmmm...
Combi Geometry is accepted...
by pro_4_ever, Feb 10, 2018, 7:32 AM
Can we post Combi here?
by ayan.nmath, Feb 10, 2018, 7:31 AM
Parag dey
by Paragdey12, Feb 6, 2018, 1:17 PM
More posts please
by AnArtist, Feb 6, 2018, 2:38 AM
Second Shout!
by ccx09, Feb 3, 2018, 9:20 PM
1st shout!
by AnArtist, Jan 31, 2018, 9:20 AM

23 shouts

Problems In Geometry

#4) Symmedians in a Triangle

by AnArtist, Feb 7, 2018, 4:32 PM

4 Comments